Problem: Multiply the following complex numbers: $({-4-2i}) \cdot ({-3+4i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-4-2i}) \cdot ({-3+4i}) = $ $ ({-4} \cdot {-3}) + ({-4} \cdot {4}i) + ({-2}i \cdot {-3}) + ({-2}i \cdot {4}i) $ Then simplify the terms: $ (12) + (-16i) + (6i) + (-8 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 12 + (-16 + 6)i - 8i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 12 + (-16 + 6)i - (-8) $ The result is simplified: $ (12 + 8) + (-10i) = 20-10i $